The process of cellular respiration causes a(n)

11. If there are 8.24 x 1022 molecules of NaCl in a salt shaker, what is the mass of the salt?

vlada-n [284]

Answer: 8.12 g NaCl

Explanation: Use Avogadro's number to find the number of m

moles of NaCl:

8.24x10²² molecules NaCl / 1 mole NaCl/ 6.022x10²³ molecules NaCl

= 0.14 mole NaCl

Next convert moles to grams NaCl using its molar mass;

0.14 mole NaCl x 58g NaCl / 1 mole NaCl

= 8.12 g NaCl

6 0

3 years ago

How to convert volts to electron volts?

SOVA2 [1]

Answer:

How to convert volts to electron-volts

How to convert electrical voltage in volts (V) to energy in electron-volts (eV).

You can calculate electron-volts from volts and elementary charge or coulombs, but you can't convert volts to electron-volts since volt and electron-volt units represent different quantities.

Volts to eV calculation with elementary charge

The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electric charge Q in elementary charge or proton/electron charge (e):

E(eV) = V(V) × Q(e)

The elementary charge is the electric charge of 1 electron with the e symbol.

So

electronvolt = volt × elementary charge

or

eV = V × e

Example

What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 40 electron charges?

E = 20V × 40e = 800eV

Volts to eV calculation with coulombs

The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C) divided by 1.602176565×10-19:

E(eV) = V(V) × Q(C) / 1.602176565×10-19

So

electronvolt = volt × coulomb / 1.602176565×10-19

or

eV = V × C / 1.602176565×10-19

Example

What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 2 coulombs?

E = 20V × 2C / 1.602176565×10-19 = 2.4966×1020eV

Explanation:

4 0

3 years ago

An unknown element is determined to have 22 neutrons and a mass number of 40.

Amiraneli [1.4K]

It’s Calcium hope that helps!

7 0

3 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

Some rice farmers burn their fields after harvest to remove leftover plant material and kill certain plant pests. You are resear

WARRIOR [948]

I think its proably 3.....

5 0

3 years ago